Q: Summation of zeta-values and of bernoulli-numbers

I got two results, which I like to have confirmed:

1)

1*(zeta(2)-1) + 1*(zeta(3)-1) + 1*(zeta(4)-1) + 1*(zeta(5)-1) + ... = 1
2*(zeta(3)-1) + 3*(zeta(4)-1) + 4*(zeta(5)-1) + ... = 1
3*(zeta(4)-1) + 6*(zeta(5)-1) + ... = 1
..
// for a>0
sum(i=a,n,binomial(i,a)*(zeta(1+i)-1))-binomial(a,a)*(zeta(1+a)-1) = 1

Are these all convergent, btw?

2)
With the following divergent summation I feel insure.
I got, with b_k the k'th bernoulli-number, b_1 = +1/2

sum(k=1,inf, (1 - 2^(-k))* b_(k-1) ) = 1


with b_1 = -1/2

sum(k=1,inf, (1 - 2^(-k))* b_(k-1) ) = 1/4

Both results seem to occur by a certain matrix-inversion, but I can't
check this by, say, Borel-summation. In 2) I also have variants
involving binomial-coefficients similar to that in 1), but I think,
if already the given variant would be wrong, they should all be similarly
wrong.

TIA -

Gottfried Helms
.

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